8
$\begingroup$

Poincare duality gives us, for a smooth orientable $n$-manifold, an isomorphism $H^k(M) \to H_{n-k}(M)$ given by $\gamma \mapsto \gamma \frown [M]$ where $[M]$ is the fundamental class of the manifold, and $\frown$ is the cap product.

This allows us (for example) to, given an embedding $f : X \to Y$ of manifolds to push forward integration:

$$ \int_X f^*\omega = \int_Y \omega \smile (f_*[X])^\vee $$

which is a simple application of the push-pull formula: $f_*(a \frown f^*b) = (f_*a) \frown b$.

To what extent do results like this hold for stacks such as $\overline{M}_{g,n}(X,\beta)$? I know that these are not in general nice, that they can have components of a variety of dimensions, etc. They do, however, have virtual fundamental classes of pure dimension which behave in many ways like the fundamental classes that we would want.

My question is then the following:

Is there a Poincare duality for such stacks, using the virtual fundamental class? Does the push-pull formula hold in such a case?

Bonus question: What about moduli stacks with reduced obstruction theories, such as for studying Gromov-Witten invariants of K3 and Abelian surfaces? Is there any chance that it is still true there?

$\endgroup$
4
  • $\begingroup$ Simon -- what are the coefficients? $\endgroup$
    – algori
    Nov 24, 2011 at 1:56
  • $\begingroup$ Since I'm talking about stacks, I would be thinking rational ones. In that case. If we do so, for example, we do get poincare duality for orbifolds. $\endgroup$
    – Simon Rose
    Nov 24, 2011 at 2:28
  • 1
    $\begingroup$ I am not sure I understand the question. The virtual fundamental class could very well be 0, while the space is not empty; I don't see how you could expect a Poincaré duality. Perhaps there is a "virtual homology" for which Poincaré duality holds, but I have never seen this done. $\endgroup$
    – Angelo
    Nov 24, 2011 at 7:35
  • $\begingroup$ Angelo - I'm not really sure that it's a well-posed question either. It was perhaps a bit of a fishing expidition to see just to what extent the virtual fundamental class behaves like an ordinary fundamental class. The lack of an obvious choice of homology theory that would make sense with it, I admit, makes it unlikely. $\endgroup$
    – Simon Rose
    Nov 24, 2011 at 17:08

1 Answer 1

6
$\begingroup$

I know of one very limited case, which will almost certainly not satisfy you, where the "virtual cohomology" mentioned by Angelo exists, with a version of Poincare duality.

Presumably the generality one would like to address this question is for a moduli space (or stack) $M$ carrying a perfect obstruction theory (let me ignore the stackyness everywhere below). Let me restrict right away to virtual dimension 0, and in fact further to the case when the obstruction theory is symmetric (in the sense of Behrend, Behrend-Fantechi). Note that this already excludes Kontsevich moduli spaces $\bar M_{g,n}(X,\beta)$, even for $X$ a Calabi--Yau threefold, but we are still OK with DT- or other sheaf-theoretic moduli spaces.

In this case, we of course have Behrend's result that locally $M$ is cut out in a smooth ambient space $N$ by the zeros of an almost closed one-form. Now specialize further, and assume that in fact $M$ is cut out by the zeros of a closed one-form. Finally, specialize to the case when in fact globally $M=Z(df)\subset N$; here $N$ is a smooth variety, $f\colon N\to {\mathbb C}$ is a smooth function, and $M$ is cut out by the zeros of the exact one-form $df$. Note that there are still reasonably interesting examples, such as the Hilbert scheme of points on ${\mathbb C}^3$ or related toric or quivery examples.

In this case, it seems that the correct "virtual cohomology" to consider is $H^*(M,\phi_f)$, the cohomology of $M$ with coefficients in the perverse $\mathbb Q$-sheaf $\phi_f\in{\rm Perv}_{\mathbb Q}(M)$ of vanishing cycles of $f$ (appropriately shifted). This is called critical cohomology by Kontsevich-Soibelman. For example, this cohomology has the correct Euler characteristic, namely the integral of the Behrend function on $M$. It also carries a mixed Hodge structure, and the Euler characteristic of the weight filtration (which will differ from the degree filtration because the Hodge structure is not pure) gives what seems to be the right "quantization" (or "refinement") of the numerical invariant.

Now the point is that by standard theory, $D_M\phi_f\cong\phi_f$, where $D_M$ is the Verdier duality functor on $M$. This will give you a kind of Poincare duality $$H^n(M,\phi_f)\cong H^{-n}_c(M,\phi_f)$$ albeit between ordinary and compact-support cohomology, since $M$ is of course almost always noncompact. But at least (using the appropriate shifts) it is elegantly symmetric around zero, which is what we would expect in the virtual dimension 0 case.

I don't really know how well this generalizes; almost nothing is known (to me) about gluing, functoriality, etc.

$\endgroup$

Your Answer

By clicking “Post Your Answer”, you agree to our terms of service and acknowledge that you have read and understand our privacy policy and code of conduct.

Not the answer you're looking for? Browse other questions tagged or ask your own question.